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# A wire of fixed length is wound on a solenoid of length $l$ and radius $R$. Its Self Inductance is found to be $L$. Now, if the same wire is wound on a solenoid of length $\frac{I}{2}$ and radius $\frac{R}{2}$, then self inductance will be

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Solution

## Step 1: Given$\text{Originallength}=l$$\text{Newlength}=\frac{l}{2}$$\text{originalinductance}=L$Step 2: To findWe have to find the new inductance.Step 3: Calculate the new inductanceUse the formula for self-inductance.The expression for self-inductance is given by,$L=\frac{{\mu }_{0}{N}^{2}A}{l}...\left(1\right)$Where, ${\mu }_{0}$ is the magnetic permeability of free space, ${N}^{2}$ is the number of turns per unit length, $A$ is the area of the solenoid, and, $l$ is the length of the solenoid.The area of a solenoid is given by,$A=\mathrm{\pi }{\mathrm{r}}^{2}$Where, $\mathrm{r}$ is the radius of the cross-section of the solenoid.Substituting this value in equation (1). $L=\frac{{\mu }_{0}{N}^{2}\mathrm{\pi }{\mathrm{r}}^{2}}{l}...\left(2\right)$ Now, the length of the wire is given by,$\mathrm{length}=\mathrm{circumference}×\mathrm{number}\mathrm{of}\mathrm{turns}$$⇒l=2\mathrm{\pi r}×N$Since the same wire is taken in both cases, then the circumference and number of turns will be the same for both wires.$⇒L=2\mathrm{\pi r}×N$ Rearranging the above equation.$r=\frac{L}{2\mathrm{\pi }N}...\left(3\right)$Now, substitute equation (3) in equation (2).$L=\frac{{\mu }_{0}{N}^{2}\mathrm{\pi }{\left(\frac{\mathrm{L}}{2\mathrm{\pi N}}\right)}^{2}}{l}$$⇒L=\frac{{\mu }_{0}{L}^{2}}{4\pi l}$This will be the value of $L$ when the length of the wire is $l$ and radius is $\mathrm{r}$.For, length $\frac{l}{2}$ and radius $\frac{r}{2}$ value of $L$ will be,${L}_{0}=\frac{{\mu }_{0}{L}^{2}}{4\pi \frac{l}{2}}$$⇒{L}_{0}=2\frac{{\mu }_{0}{L}^{2}}{4\pi l}$Thus, the self-inductance of the coil will be doubled.Hence, the new self-inductance will be two times the original self-inductance.

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